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According to mathworld, Ramanujan's master theorem is the statement that if $$f(z) = \sum_{k=0}^{\infty} \frac{\phi(k) (-z)^k}{k!}$$ for some function (analytic or integrable) $\phi$, then $$\int_0^{\infty} x^{n-1} f(x) \, \mathrm{d}x = \Gamma(n) \phi(-n).$$

As written it is obviously false as the values of an (analytic or integrable) function $\phi$ at natural numbers do not determine its values anywhere else. However it turns out that $$\int_0^{\infty} x^{s-1} f(x) \, \mathrm{d}x = \Gamma(s) \phi(-s)$$ for arbitrary $s$ under growth conditions on $\phi$.

Recently I came across an elementary "proof": if $T$ denotes the shift operator $T\phi(s) := \phi(s+1),$ then we can write $$f(z) = \sum_{k=0}^{\infty} \frac{(-z)^kT^k \phi(0)}{k!} = e^{-zT}\phi (0)$$ such that $$\int_0^{\infty} x^{n-1} f(x) \, \mathrm{d}x = \int_0^{\infty} x^{n-1} e^{-xT} \phi(0) \, \mathrm{d}x = \Gamma(n) T^{-n}\phi(0) = \Gamma(n) \phi(-n),$$ by plugging $T$ into the Gamma integral $$\int_0^{\infty} x^{n-1} e^{-xs} \, \mathrm{d}x = \Gamma(n) s^{-n}.$$ I am curious whether this argument can be made rigorous with functional analysis on an appropriate function space (which necessarily would have to have some growth conditions).

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    $\begingroup$ I remember the first time I saw this kind of heuristic derivation, it was with Euler–Maclaurin formula. here too it's enlightening. I think Berndt gave a 'rigourous proof' of Ramanujan's master theorem. $\endgroup$ Jan 1, 2017 at 8:33
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    $\begingroup$ I don't think it is "obviously false as written". Because $f$ is uniquely determined by the sequence $(\psi (k))_k.$ Any sequence $(A_k)_k$ is equal to $(\psi (k))_k$ for some $\psi $ integrable from $0$ to $\infty.$ Some restrictions needed are that the radius of convergence of the series for $f$ is $\infty$, and that $x^{n-1}f(x)$ is integrable from $0$ to $\infty$ for every positive integer $n.$ $\endgroup$ Jan 1, 2017 at 10:23
  • $\begingroup$ @user254665 The point is that $\psi$ is not uniquely determined by $f$ $\endgroup$
    – user6246
    Jan 1, 2017 at 16:50
  • $\begingroup$ I think it is uniquely determined by $f$. $\endgroup$ Nov 4, 2018 at 10:53
  • $\begingroup$ The formula of Ramanujan is based on Mellin transform and a proof was given by G H Hardy in Ramanujan : Twelve Lectures on Subjects Suggested by His Life and Work. See page $189$. $\endgroup$
    – Paramanand Singh
    Nov 5, 2018 at 9:00

1 Answer 1

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Write $$ (M\Psi)(s)=\int^{\infty}_{0}\Psi(t)t^{s-1}dt, $$ where $s\in A_{\Psi}$ with $$ A_{\Psi}:=\left\{s\in\textbf{C}:\int^{\infty}_{0}\Psi(t)t^{s-1}dt<\infty\right\}. $$ Then we have the next Mellin inversion formula $$ \Psi(z)=\frac{1}{2\pi i}\int^{\sigma+i\infty}_{\sigma-i\infty}(M\Psi)(s)z^{-s}ds, $$ where $\sigma\in Re\left(A_{\Psi}\right)$.

Theorem 1. If $\Psi(z)$ have power series arround $0$ with radious of convergence $r>0$ and if $x\in\textbf{R}$ such that $$ \int^{\infty}_{0}\left|\Psi(t)\right|t^{x-1}dt<+\infty. $$ Then the Mellin transform of $\Psi$ can be analyticaly continued to a meromorphic function in the halphplane $Re(z)<x$, with poles at the points $z=-m$, $m$ is non-negative integers such $m>-x$.

Theorem 2. Let $x>0$ and $f,\Psi$ analytic in $\textbf{C}$ and satisfining the condition $$ |f(z)(M\Psi)(x+iz)|\leq C (1+|z|)^{\lambda}e^{-\delta |Re(z)|},\tag 1 $$ for all $z$ such that $Im(z)\geq 0$ and $c,\lambda,\delta>0$ constants, with the condition that $|z|=x+N+1/2$, where $N-$natural number saficiently large. Then the integral $$ \int^{\infty}_{-\infty}f(t)M\Psi(x+it)dt $$ converges absolutely, the series $$ \sum^{\infty}_{m=0}\frac{\Psi^{(m)}(0)}{m!}f(i(x+m)) $$ converges in Abel sence and $$ \int^{\infty}_{-\infty}f(t)(M\Psi)(x+it)dt=2\pi\lim_{r\rightarrow 1}\sum^{\infty}_{m=0}\frac{\Psi^{(m)}(0)}{m!}f(i(x+m))r^m.\tag 2 $$ Moreover if $$ \left|\frac{\Psi^{(m)}(0)}{m!}f(i(x+m))\right|\leq \frac{C'}{m+1}, $$ then (2) converges.

As application of the above theorem we have

Theorem 3. Let $f$ be analytic in the upper half plane $Im(z)>0$ and continuous in $Im(z)\geq 0$ and such that $$ |f(z)|\leq C (1+|z|)^{\rho}\left(\frac{|z|}{e}\right)^{Im(z)}e^{b|Re(z)|}, $$ in $Im(z)\geq 0$, where $0\leq b\leq \pi/2$. Then for $x>0$ we have $$ \int^{\infty}_{-\infty}f(t)\Gamma(x+it)dt=2\pi\lim_{r\rightarrow1}\sum^{\infty}_{m=0}\frac{(-1)^m}{m!}f(i(x+m))r^m $$

Theorem 3 is Ramanujan's Master Theorem.

I have full proofs of that in my PhD thesis but there are written in Greek (see Researchgate: Nikos Bagis).

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  • $\begingroup$ Sir. I looked up your research gate but could not find the paper you mention. A reference to a rigorous derivation of Ramanujan’s theorem would be amazing! $\endgroup$
    – FShrike
    Dec 4, 2021 at 21:32
  • $\begingroup$ @FShrike. See at researchgate.net/publication/235945980_Thesis_NBagis $\endgroup$ Dec 6, 2021 at 6:28
  • $\begingroup$ Many thanks, I’ll have a go at reading this $\endgroup$
    – FShrike
    Dec 6, 2021 at 6:49

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